1)Lin solved the equation incorrectly. Find all the errors in her solution. (Select all appropriate answers.)Given to Step 1: Lin calculated 4 - 17 incorrectly; should be -13.Step 1 to Step 2: Lin distributed 8(x - 3) incorrectly; should be 8x - 3.Step 2 to Step 3: Lin calculated 8x - 24 + 7 incorrectly; should be 8x - 31.Step 3 to Step 4: Lin should have subtracted 8x from each side; should be 18x.What should Lin's answer have been ?8 ( x - 3 ) + 7 = 2x ( 4 - 17 )answer choices :x = 3/2 x = 1/2x = 1/5x = 1/3
Answer:
Given to Step 1: Lin calculated 4 - 17 incorrectly; should be -13.
Step 3 to Step 4: Lin should have subtracted 8x from each side; should be 18x.
Lin's answer should be x = 1/2
Explanation:
The initial expression is:
8(x - 3) + 7 = 2x(4 - 17)
So, the first error is in step 1, because 4 - 17 = - 13, then step 1 should be:
8(x - 3) + 7 = 2x(-13)
Then, we can apply the distributive property as:
8x - 8(3) + 7 = -26x
8x - 24 + 7 = -26x
We need to add similar terms:
8x - 17 = - 26x
Then, we can subtract 8x from both sides as follows:
8x - 17 - 8x = - 26x - 8x
-17 = -34x
Finally, we can divide by -34 as follows:
-17/(-34) = -34x/(-34)
1/2 = x
Therefore, the answers are:
Given to Step 1: Lin calculated 4 - 17 incorrectly; should be -13.
Step 3 to Step 4: Lin should have subtracted 8x from each side; should be 18x.
Lin's answer should be x = 1/2
Can you please help me
Step 1
A rectangle is a quadrilateral with the parallel opposite two sides being equal and with each vertex being 90°. The sum of angles in a rectangle is 360°. It has 2 diagonals.
Step 2
Derive an answer from the statement in step 1
The consecutive sides are perpendicular is the only right option.
Hence option D
What is the length of BC to the nearest 10th of a centimeter
Step 1: sketch out the right angled triangle
Step 2: calculate the value BC
To calculate the value of BC we will use the trigonometric ratio
[tex]\begin{gathered} \sin \theta=\frac{opp}{hyp} \\ \text{where,} \\ \text{opp}=BC \\ \text{Hyp}=6.2\operatorname{cm} \\ \theta=32^0 \end{gathered}[/tex]By substituting the values, we will have
[tex]\begin{gathered} \sin 32^0=\frac{BC}{6.2} \\ BC=6.2\times\sin 32^0 \\ BC=6.2\times0.5299 \\ BC=3.2854\operatorname{cm} \\ \text{appro}\xi\text{mately to the nearest tenth we will have} \\ BC=3.3\operatorname{cm} \end{gathered}[/tex]Hence,
The value of BC =3.3cm
Lauren uses1/3cup of carrot juice for every2/3cup of apple juice to make a fruit drink.Enter the number of cups of carrot juice Lauren uses for 1 cup of apple juice,
1/3 cup of carrot juice ---------------------------------->2/3 cup of apple juice
x cups of carrot juice ----------------------------------->1 cup of apple juice
Using cross multiplication:
[tex]\frac{\frac{1}{3}}{x}=\frac{\frac{2}{3}}{1}[/tex]solve for x:
[tex]\begin{gathered} x=\frac{\frac{1}{3}}{\frac{2}{3}} \\ x=\frac{3}{6} \\ x=\frac{1}{2} \end{gathered}[/tex]She uses 1/2 cups of carrot juice for 1 cup of apple juice
Determine whether the Mean Value theorem can be applied to f on the closed interval [a, b]. (Select all that apply.)
Solution
The given function is
[tex]f(x)=4x^3[/tex]With given interval
[tex]\lbrack1,2\rbrack[/tex]The function is differentiable on the open interval (1,2) and it is continuous on the closed interval [1,2]
Therefore mean value theorem can be used
Calculating the c value iit follows:
[tex]f^{\prime}(c)=\frac{f(2)-f(1)}{2-1}[/tex]This gives
[tex]\begin{gathered} f^{\prime}(c)=\frac{4(2)^3-4(1)^3}{1} \\ f^{\prime}(c)=\frac{32-4}{1} \\ f^{\prime}(c)=28 \end{gathered}[/tex]Differentiating the given function gives:
[tex]f^{\prime}(x)=12x^2[/tex]Equate f'(c) and f'(x)
This gives
[tex]x^2=\frac{28}{12}[/tex]Solve the equation for x
[tex]\begin{gathered} x^2=\frac{28}{12} \\ x^2=\frac{7}{3} \\ x=\pm\sqrt{\frac{7}{3}} \\ x=\sqrt{\frac{7}{3}},x=-\sqrt{\frac{7}{3}} \end{gathered}[/tex]Therefore the values of c are
[tex]\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}[/tex]Can you please help me with 28 Please give all end behavior such as limits and as_,_
We must describe the local and end behaviour of the function:
[tex]f(x)=\frac{x^2-4x+3}{x^2-4x-5}.[/tex]First, we rewrite the polynomials in numerator and denominator in terms of their roots:
[tex]f(x)=\frac{(x-1)\cdot(x-3)}{(x+1)\cdot\mleft(x-5\mright)}\text{.}[/tex]Local behaviour
We see that f(x) has a zero in the denominator for x = -1 and x = 5. The function f(x) has vertical asymptotes at these values. To analyze the local behaviour, we must compute the lateral limits for x → -1 and x → 5.
Limit x → - 1 from the left
Computing the limit from the left when x → -1, is equivalent to replacing x by -1 - ε and computing the limit when ε → 0:
[tex]\begin{gathered} \lim _{x\rightarrow-1^-}f(x)=\lim _{\epsilon\rightarrow0}f(-1-\epsilon) \\ =\lim _{\epsilon\rightarrow0}\frac{(-1-\epsilon-1)\cdot(1-\epsilon-3)}{(-1-\epsilon+1)\cdot(-1-\epsilon-5)} \\ =\lim _{\epsilon\rightarrow0}\frac{(-2)\cdot(-2)}{(-\epsilon)\cdot(-6)}\rightarrow+\infty. \end{gathered}[/tex]In the last step, we can't throw the ε in the parenthesis different to zero.
Limit x → - 1 from the right
Computing the limit from the left when x → -1, is equivalent to replacing x by -1 + ε and computing the limit when ε → 0:
[tex]\begin{gathered} \lim _{x\rightarrow-1^+}f(x)=\lim _{\epsilon\rightarrow0}f(-1+\epsilon) \\ =\lim _{\epsilon\rightarrow0}\frac{(-1+\epsilon-1)\cdot(1+\epsilon-3)}{(-1+\epsilon+1)\cdot(-1+\epsilon-5)} \\ =\lim _{\epsilon\rightarrow0}\frac{(-2)\cdot(-2)}{(+\epsilon)\cdot(-6)}\rightarrow-\infty. \end{gathered}[/tex]In the last step, we can't throw the ε in the parenthesis different to zero.
Limit x → 5 from the left
Computing the limit from the left when x → 5, is equivalent to replacing x by 5 - ε and computing the limit when ε → 0:
[tex]\begin{gathered} \lim _{x\rightarrow-1^-}f(x)=\lim _{\epsilon\rightarrow0}f(-1-\epsilon) \\ =\lim _{\epsilon\rightarrow0}\frac{(5-\epsilon-1)\cdot(5-\epsilon-3)}{(5-\epsilon+1)\cdot(5-\epsilon-5)} \\ =\lim _{\epsilon\rightarrow0}\frac{(+4)\cdot(+2)}{(+4)\cdot(-\epsilon)}\rightarrow-\infty. \end{gathered}[/tex]In the last step, we can't throw the ε in the parenthesis different to zero.
Limit x → 5 from the right
Computing the limit from the left when x → 5, is equivalent to replacing x by 5 + ε and computing the limit when ε → 0:
[tex]\begin{gathered} \lim _{x\rightarrow-1^-}f(x)=\lim _{\epsilon\rightarrow0}f(-1-\epsilon) \\ =\lim _{\epsilon\rightarrow0}\frac{(5+\epsilon-1)\cdot(5+\epsilon-3)}{(5+\epsilon+1)\cdot(5+\epsilon-5)} \\ =\lim _{\epsilon\rightarrow0}\frac{(+4)\cdot(+2)}{(+4)\cdot(+\epsilon)}\rightarrow+\infty. \end{gathered}[/tex]In the last step, we can't throw the ε in the parenthesis different to zero.
End behaviour
To describe the end behaviour of the function, we must compute the limits of the function when x → -∞ and x → +∞.
Limit x → -∞
[tex]\begin{gathered} \lim _{x\rightarrow-\infty^{}}f(x)=\lim _{x\rightarrow-\infty^{}}\frac{x^2-4x+3}{x^2-4x-5} \\ =\lim _{x\rightarrow-\infty}\frac{x^2-4x+3}{x^2-4x-5}=\frac{\lim _{x\rightarrow-\infty}\frac{x^2-4x+3}{x^2}}{\lim _{x\rightarrow-\infty}\frac{x^2-4x-5}{x^2}}=\frac{1}{1}=1. \end{gathered}[/tex]To compute the limit we have divided numerator and denominator by x² and distributed the limit. The result of each limit is given by the leading term, which has the highest power of x.
Limit x → +∞
[tex]\begin{gathered} \lim _{x\rightarrow+\infty^{}}f(x)=\lim _{x\rightarrow+\infty^{}}\frac{x^2-4x+3}{x^2-4x-5} \\ =\lim _{x\rightarrow+\infty}\frac{x^2-4x+3}{x^2-4x-5}=\frac{\lim_{x\rightarrow+\infty}\frac{x^2-4x+3}{x^2}}{\lim_{x\rightarrow+\infty}\frac{x^2-4x-5}{x^2}}=\frac{1}{1}=1. \end{gathered}[/tex]To compute the limit we have divided numerator and denominator by x² and distributed the limit. The result of each limit is given by the leading term, which has the highest power of x.
AnswersLocal behaviour
The function f(x) has vertical asymptotes at x = -1 and x = 5.
[tex]\begin{gathered} \lim _{x\rightarrow-1^-}f(x)=+\infty \\ \lim _{x\rightarrow-1^+}f(x)=-\infty \\ \lim _{x\rightarrow5^-}f(x)=-\infty \\ \lim _{x\rightarrow5^+}f(x)=+\infty \end{gathered}[/tex]End behaviour
[tex]\begin{gathered} \lim _{x\rightarrow-\infty^{}}f(x)=1 \\ \lim _{x\rightarrow+\infty^{}}f(x)=1 \end{gathered}[/tex]Which of the statements below is true for the following set of numbers? 20, 15, 50, 85, 75, 60A. The range is 70 and the midrange is 35 B.the range is 70 and the midrange is 50C. The range is 85 and the midrange is 55D.the range and the midrange are equal
B
1) Firstly, we need to orderly write this sequence of numbers:
[tex]15,20,50,60,75,85[/tex]2) Then we need to calculate the Range of this Data Set, by subtracting from the highest values the least one:
[tex]R=85-15=70[/tex]2.2) The Midrange is the average between the Highest Value and the Least value on this Data Set:
[tex]M=\frac{85+15}{2}=50[/tex]3) Thus the answer is B
write the er exponent expression forty-one to the seventh power
Write the exponential expression, forty one to the seventh power;
[tex]41^7[/tex]Without graphing, identify the vertex, axis of symmetry, and transformations from the parent function f(x)= |x|y = |x-5|-1
Answer:
• Vertex: (5, –1)
,• No symmetry
,• Transformations: 5 units to the right, and 1 unit down.
Explanation
We are given the parent function f(x)= |x| and the transformed function:
[tex]y=|x-5|-1[/tex]Thus, we can get the vertex considering that a function in the form:
[tex]y=|x\pm a|\pm b[/tex]has a vertex at (+a, ±b).
Therefore, our vertex is at (5, –1). Additionally, as an absolute function has the form of a 'v', and as the vertex is at (5, –1) then it has no symmetry about the x-axis, nor y-axis, and nor about the origin, meaning it has no symmetry.
Finally, the transformation from the parent function is a shift 5 units to the right and one unit down.
Answer:
Answer:
• Vertex: (5, –1)
,
• No symmetry
,
• Transformations: 5 units to the right and 1 unit down.
Explanation
We are given the parent function f(x)= |x| and the transformed function:
Thus, we can get the vertex considering that a function in the form:
Has a vertex at (+a, ±b).
Therefore, our vertex is at (5, –1). Additionally, as an absolute function has the form of a 'v', and as the vertex is at (5, –1) then it has no symmetry about the x-axis, nor y-axis, nor about the origin, meaning it has no symmetry.
Finally, the transformation from the parent function shifts 5 units to the right and one unit down.
Step-by-step explanation:
If you were to solve the following system by substitution, what would be the best variable to solve for and from what equation? 3x + 6v=9 2x – 10v=13
The best variable to solve is x=3-2v, after dividing the first equation by 3.
x=4, and v=-1/2
1) Solving that system by Substitution
2) Making then
x=3-2v
2x-10v=13
3) Plugging into the 2nd equation
2(3-2v)-10v=13
6-4v -10v=13
6-14v=13
-14v=13-6
-14v=7
v=-1/2
Plugging into the first equation
x=3-2v
x=3-2(-1/2)
x=3+1
x=4
The table shows the predicted growth of bacteria after various numbers of hours. Write an explicit formula for the number of bacteria after n hours.Hours1 2 3 4 5(n)Numberof 33 57 81 105 129BacteriaΟ Α.a= 24+9O B. a = 9n+ 24OC. a, = 24n+33OD. a = 9n +33
Let a be the number of bacteria after one hour.
From the table, we get a=33.
The difference between 57 and 33 is 57-33= 24.
The difference between 81 and 57 is 81-57=24
Hence we get the common difference d=24.
The given data is in the arithmetic progression.
The formula for the nth term in the arithmetic progression is
[tex]a_n=a+(n-1)d[/tex]Substitute a=33 and d=24, we get
[tex]a_n=33+(n-1)24[/tex][tex]a_n=33+24n-24[/tex][tex]a_n=24n+9[/tex]Hence the required recursive equation is
[tex]a_n=24n+9[/tex]Option A is correct.
BC and DE are chords of circle A, and BC DE. Which statement cannot be verifiedfrom the information that is given?BC DEBZBACZ ZDAEAABC = AADEZDAE CAD
Let's start in the first information:
[tex]arcBC\cong arcDE[/tex]This can be verified, because, AC, AB, AD and AE are all congruent, because they are the radius of A and, since BC and DE are congruent, triangles ABC and ADE are also congruent.
Thus, the angles mBAC and mDAE are congruent. The measure of thesee angles are the same as the arcs BC and DE, so we verified this alternative.
In the explanation above, we also verified the second alternative:
[tex]\angle BAC\cong\angle DAE[/tex]And we verified the third alternative:
[tex]\Delta ABC\cong\Delta ADE[/tex]We are left with the last alternative:
[tex]\angle DAE\cong\angle CAD[/tex]This can't be verified. One way of seeing that is that we can rotate triangle ABC around the circle without changing any of the given information, however this changes mCAD.
So the only statement that cannot be verified from the given information is the last one:
[tex]\angle DAE\cong\angle CAD[/tex]Graph y-12=4(x-(-3))
The graph of the equation can be given as
What is an equation of the line how to graph the equation?
An equation is an expression between two variables With an equality sign. Usually an equation of the line can be given as y= mx + c, Where x is independent and y is dependent and c is y-intercept. as the value of x changes the value of y also changes.
We are given an equation y-12 = 4(x-(-3))
We first simplify the equation
we get y -12 = 4(x+3)
Multiplying 4 inside the bracket we get,
y-12= 4x+12
Adding 12 on both sides to get slope intercept form.
y=4x+24
The slope of the line is 4 and y-intercept is 24.
Now we can plot the given equation.
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48. Mrs. Dalton is selling pieces of cake at the school carnival. She baked 8 cakes and cut them
each into 12 pieces. If she sold 81 slices of cake, how many total cakes does she have left?
Circle the correct option.
Total number of cakes left is 1.25 i.e. 1 and quarter.
Given,
Number of cakes = 8
Number of pieces of each cake =12
Number of slices sold = 81
Then,
Total number of slices of cake = [tex]12*8=96[/tex]
Now,
number of remaining slices =[tex]96-81=15[/tex]
To find number of remaining cake left, divide number of remaining slices to the pieces of each cake.
Number of remaining cake = [tex]\frac{15}{12} =1.25[/tex]
Thus, total number of cakes left is 1.25 i.e. 1 and quarter.
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Find the first five terms of the sequence
defined recursively as follows: t₁ = 1,
tn = 3(t(n-1)), n≠1, n is a natural number.
Answer:
1, 3, 9, 27, 81
Step-by-step explanation:
using the recursive rule
[tex]t_{n}[/tex] = 3 [tex]t_{n-1}[/tex] with t₁ = 1 , then
t₂ = 3 × t₁ = 3 × 1 = 3
t₃ = 3 × t₂ = 3 × 3 = 9
t₄ = 3 × t₃ = 3 × 9 = 27
t₅ = 3 × t₄ = 3 × 27 = 81
the first 5 terms are 1, 3, 9, 27, 81
can anyone give me an example of plotting points with a real life word problem
Let's have an example of a seller that sells phones.
His salary have a fixed part of $50, and it increases by $2 by each phone he sells.
We want to find out how much will be his salary if he sells 5, 10 and 15 phones.
If we use the variable x to represent the number of phones sold, his salary (variable y) will be defined by the equation:
[tex]y=50+2x[/tex]Now, using the values of x we want to calculate, we have that:
[tex]\begin{gathered} x=5\colon \\ y=50+2\cdot5=50+10=60 \\ \\ x=10\colon \\ y=50+2\cdot10=50+20=70 \\ \\ x=15\colon \\ y=50+2\cdot15=50+30=80 \end{gathered}[/tex]We can plot these points to see how much his salary increases for each phone sold, and also we can have an idea of any point we want to find out:
-4 5/9 + (-1 2/3)[tex] - 4 \frac{5}{9} + ( - 1 \frac{2}{3} )[/tex]
Answer:
-6 2/9
Explanation:
Given the expression:
[tex]-4\frac{5}{9}+(-1\frac{2}{3})[/tex]First, open the brackets:
[tex]=-4\frac{5}{9}-1\frac{2}{3}[/tex]Next, change the fractions to improper fractions:
[tex]=-\frac{41}{9}-\frac{5}{3}[/tex]Then, take the lowest common multiple of 9 and 3 to combine the fractions:
[tex]\begin{gathered} =\frac{-41-5(3)}{9} \\ =\frac{-41-15}{9} \\ =\frac{-56}{9} \\ =-6\frac{2}{9} \end{gathered}[/tex]The result is -6 2/9.
There can be no more than 100 people in the movie theater. There are already 22 people in the movie theater.What inequality represents the number of additional people, p, that can enter the movie theater?Drag and drop the appropriate symbols to correctly complete the inequality.
Answer:
[tex]p\text{ + 22}\leq\text{ 100}[/tex]Explanation:
Here, we want to drop the appropriate symbols
When we add the given number 22 to p, it would give a number which is at most 100
That means the number must be less than or equal to 100
mathematically, we have that as:
[tex]p\text{ + 22}\leq\text{ 100}[/tex]List each real zero of f according to the behavior of the graph at the x-axis near that zero. Zero(s) where the graph crosses the x-axis:Zero(s) where the graph touches, but does not cross the x-axis:
In the graph, there are 2 zeros, one for each type.
The zero where the graph crosses the x axis is 1. When x=1, the function intercepts the x axis and crosses it.
The zero where the graph touches but does not cross the x axis is -1. When x=-1, the function touches x axis but goes back to the quadrant.
14. Log(x)=2meansa. x=10b. x=2^10c. x=10^2d. x=0e. none of the above
For a given geometric sequence, the common ratio, r, is equal to 5, and the 7th term, an, is equal to -43. Find the value of the 9thterm, a9. If applicable, write your answer as a fraction.a9=
Given:
it is given that common ration of a geometric sequence is r = 5 and 7th term is - 43.
Find:
we have to find the value of 9th term.
Explanation:
we know the formula for nth term of a geometric sequence is
[tex]a_n=ar^{n-1}[/tex]since, 7the term is - 43,
Therefore, we have
[tex]\begin{gathered} a_7=-43 \\ ar^{7-1}=-43 \\ ar^6=-43 \\ a(5)^6=-43 \\ a(15625)=-43 \\ a=-\frac{43}{15625} \end{gathered}[/tex]The 9the term of the geometric sequence is
[tex]\begin{gathered} a_9=-\frac{43}{15625}\times(5)^{9-1} \\ =-\frac{43}{15625}\times(5)^8 \\ =-\frac{43}{(5)^6}\times(5)^8 \\ =-43\times25 \\ a_9=-1075 \end{gathered}[/tex]Therefore, 9th term of given geometric sequence is -1075.
If anyone can answer this question I will be surprised
A reflection through x-axis is the property to get a point and transform it like this:
[tex]\begin{gathered} P\rightarrow P^{\prime} \\ (x,y)\rightarrow(x,-y) \end{gathered}[/tex]If we have point P = (3,4), its reflexion through x-axis will be:
[tex]\begin{gathered} P\rightarrow P^{\prime} \\ (x,y)\rightarrow(x,-y) \\ (3,4)\rightarrow(3,-4) \end{gathered}[/tex]So our final answer will be:
[tex](3,-4)[/tex]What is the location of the vertex on the parabola defined by f(x) = 2x2 + 14x + 11 and in what direction does the parabola open?
Vertex of a parabola
The point where the parabola intersects its axis of symmetry is called the "vertex" and is the point where the parabola is most sharply curved.
The given function is,
[tex]f(x)=2x^2+14x+11[/tex]We will apply graphical method to obtain the solution to the vertex of the parabola
From the graph above, the vertex of the parabola is (- 3.5, - 13.5) which can also be represented into fraction as
[tex](-\frac{7}{2},-\frac{27}{2})[/tex]And finally, the parabola opens upward.
Hence, the correct option is Option 2.
Entrance Ticket on Translations Sarah graphed a triangle with vertices X (3,3) Y (4,1) Z (1,1). She asked her classmate Paul to translate the triangle (x-4) (y+2). Paul stated that the triangle will move down by 4 and right by 2 putting the triangle in the 4th quadrant. Graph the translation to see if Paul is correct? Explain your reasoning. Your explanation should include: • What is a translation? • Is Paul correct? Why or why not • Which direction should you move the triangle? • Which quadrant is the translation located?
The coordinates of the triangle XYZ are:
X(3,3)
Y(4,1)
Z(1,1)
The translation to perform is (x-4) and (y+2)
Translations are rigid transformations of a figure, this means that the triangle will move but won't change its shape or size.
A translation over the x axis results in a horizontal movement.
If you subtract a factor k x-coordinate, the movement will be to the left.
If you add the factor k x-coordinate, the movement will be to the rigth.
In this example the translation over the x-coordinate is (x-4) → 4 unit are being subtracted to the x-coordinate of each point, this results in a horizontal movement 4 units to the left
A translation over the y axis results in a vertical movement.
If you subtract a factor m from the y-coordinate, the movement will be downwards.
If you add the factor m from the y-coordinate, the movement will be upwards
In this example the translation over the y-coordinate is (y+2)→ 2 units were added to the y-coordinate, this results in a vertical movement 2 units up.
Paul moved the triangle 4 units down and 2 units right, he performed the wrong translation.
THe triangle was moved "left" and "up", the translation will be located in the second quadrant
Write the following ratio as a fraction in lowest terms. 36 to 38
Given:
Write the following ratio as a fraction in the lowest terms.
36 to 38
the ratio will be as follows:
[tex]36\text{ }to\text{ }38=\frac{36}{38}=\frac{2*18}{2*19}=\frac{18}{19}[/tex]So, the answer will be: 18/19
HiThe scatter plot shows a hiker's elevation above sea level during a hike from the base to the
top of a mountain. The equation of a trend line for the hiker's elevation is y = 7.96x +676, where x
represents the number of minutes and y represents the hiker's elevation in feet. Use the equation of
the trend line to estimate the hiker's elevation after 170 minutes.
Answer:
2029.2 ft
Step-by-step explanation:
If x represents the number of minutes, then all you have to do is plug in 170 ( the number of minutes ) into the equation.
y = 7.96 (170) + 676
y = 1353.2 + 676
y = 2029.2
y represents the hiker's elevation in feet, so the answer would be 2029.2 ft.
let me know if anything is confusing :-))
What is the equation of a line that passes through the point (8, -4) and is parrallel to the line 6x + 2y = 9
The equation of a line that passes through the point (8, -4) and is parallel to the line 6x + 2y = 9 is 3x+y-20=0.
6x + 2y - 9 = 0
2y = -6x + 9
divide by 2 on both sides
y = -3x + 9/2
slope m = -3 and c = 9/2.
Line equation through point(8, -4) is y-y1 = m(x-x1)
y - (-4) = -3(x-8)
y+4=-3x+24
3x+y-20=0.
Therefore the equation of a line that passes through the point (8, -4) and is parallel to the line 6x + 2y = 9 is 3x+y-20=0.
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Question 1.) d.) how to find instantaneous rate of change if x= 2pi
Explanation
We are given the following function:
[tex]f(x)=3\cos2x[/tex]We are required to determine the instantaneous rate of change at x = 2π.
This is achieved thus:
[tex]\begin{gathered} f(x)=3\cos2x \\ \frac{\triangle f(x)}{\triangle x}=3\cdot-2\sin2x \\ \frac{\operatorname{\triangle}f(x)}{\operatorname{\triangle}x}=-6\sin2x \\ \text{ At the point }x=2\pi \\ \frac{\operatorname{\triangle}f(x)}{\operatorname{\triangle}x}=0 \end{gathered}[/tex]Hence, the answer is:
[tex]\frac{\operatorname{\triangle}f(x)}{\operatorname{\triangle}x}=0[/tex]Write an equation for the line parallel to g(x)= -2x-6 and passing through the point (7, 4) Write the answer in slop intercept form.
The slope-intercept form is:
[tex]y=mx+c[/tex]So first we will find the gradient:
Parallel lines have the same gradient:
[tex]\begin{gathered} g(x)=-2x-6 \\ \text{The gradient from the equation above is -2} \end{gathered}[/tex]So now that we know the gradient of the line as -2, we will then find the equation of the line using the formula below as it passes through (7,4):
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \text{From (7,4)} \\ x_1=7,y_1=4 \\ y-4=-2(x-7) \\ y-4=-2x+14 \\ y=-2x+14+4 \\ y=-2x+18 \end{gathered}[/tex]Find the exact coordinates of the HOLE in this rational function: R (x)=×+1/×+(-1)